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Math Help - Am I missing a step? quoitent rule

  1. #1
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    Question Am I missing a step? quoitent rule

    I have to differentiate the following:

    f(x)=\frac{2}{ 3x-2 }

    Alright simple enough so i follow the quotient rule for this question:

    f(x)=\frac{g(x)\cdot f'(x) - f(x)\cdot (g'(x)}{[g(x)]^2 }

    \frac{(3x-2)(0)-(2)(3)}{(3x-2)^2 }
    \frac{(3x-2)-6}{(3x-2)^2 }

    When i check the answer in mathway it says it should be :

    -\frac{6}{(3x-2)^2 }


    So not really sure about what happens to the 3x-2? is the question still right if you only get to:

    \frac{(3x-2)-6}{(3x-2)^2 }
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  2. #2
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    Quote Originally Posted by sara213 View Post
    I have to differentiate the following:

    f(x)=\frac{2}{ 3x-2 }

    Alright simple enough so i follow the quotient rule for this question:

    f(x)=\frac{g(x)\cdot f'(x) - f(x)\cdot (g'(x)}{[g(x)]^2 }

    \frac{(3x-2)(0)-(2)(3)}{(3x-2)^2 }
    What is 0 times (3x-2)?

    \frac{(3x-2)-6}{(3x-2)^2 }

    When i check the answer in mathway it says it should be :

    -\frac{6}{(3x-2)^2 }


    So not really sure about what happens to the 3x-2? is the question still right if you only get to:

    \frac{(3x-2)-6}{(3x-2)^2 }
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by sara213 View Post
    I have to differentiate the following:

    f(x)=\frac{2}{ 3x-2 }

    Alright simple enough so i follow the quotient rule for this question:

    f(x)=\frac{g(x)\cdot f'(x) - f(x)\cdot (g'(x)}{[g(x)]^2 }

    \frac{(3x-2)(0)-(2)(3)}{(3x-2)^2 }
    \frac{(3x-2)-6}{(3x-2)^2 }

    When i check the answer in mathway it says it should be :

    -\frac{6}{(3x-2)^2 }


    So not really sure about what happens to the 3x-2? is the question still right if you only get to:

    \frac{(3x-2)-6}{(3x-2)^2 }
    I'd use the chain rule and exponent laws rather than quotient rule

    f(x) = 2(3x-2)^{-1}

    f'(x) = 2 \cdot \dfrac{d}{dx}(3x-2)^{-1} \cdot \dfrac{d}{dx}(3x) = \dfrac{6}{(3x-2)^2}
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    I'd use the chain rule and exponent laws rather than quotient rule

    f(x) = 2(3x-2)^{-1}

    f'(x) = 2 \cdot \dfrac{d}{dx}(3x-2)^{-1} \cdot \dfrac{d}{dx}(3x) = \dfrac{6}{(3x-2)^2}
    It's six of one, a half-dozen of the other. I find that the quotient rule, while a little more work up front, can save algebra later, since you don't have to get common denominators and add fractions (in case the numerator is more complicated than in the present problem).
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